Problem: $\overline{AC}$ is $8$ units long $\overline{BC}$ is $10$ units long $\overline{AB}$ is $2\sqrt{41}$ units long What is $\cos(\angle ABC)$ ? $A$ $C$ $B$ $8$ $10$ $2\sqrt{41}$
Answer: SOH CAH TOA os = djacent over ypotenuse adjacent $= \overline{BC} = 10$ hypotenuse $= \overline{AB} = 2\sqrt{41}$ $\cos(\angle ABC )=\frac{10}{2\sqrt{41}}$ $=\dfrac{5\sqrt{41} }{41}$